優化循環。更快的ResultList.append（[c，d，c [1]/d [1]]）？陣列？地圖？

Question

以下效果很好，但我想讓它更快。實際的應用程序可以處理Tuple1和Tuple2，每個元素有30,000個元素和17個嵌套序列。我看到很多關於更快循環的問題，並且嘗試了一個沒有改進的數組和映射。

for i in Tuple1: 
print i 

((1, 2.2, 3), (2, 3.3, 4)) 
((5, 6.6, 7), (6, 7.7, 8)) 

for i in Tuple2: 
print i 

((10, 11, 12), (11, 12, 13), (12, 13, 14)) 
((20, 21, 22), (21, 22, 23), (22, 23, 24)) 

ResultList = [] 

for a in Tuple1: 
for b in Tuple2: 
    for c in a: 
    for d in b: 
    ResultList.append([c, d, c[1]/d[1]]) 
    SomeFunction() # Processes ResultList and is not a concern. 
    ResultList=[] 

由SomeFunction處理的ResultList的示例。

[(1, 2.2, 3), (10, 11, 12), 0.2] 
[(1, 2.2, 3), (11, 12, 13), 0.18333333333333335] 
[(1, 2.2, 3), (12, 13, 14), 0.16923076923076924] 
[(2, 3.3, 4), (10, 11, 12), 0.3] 
[(2, 3.3, 4), (11, 12, 13), 0.27499999999999997] 
[(2, 3.3, 4), (12, 13, 14), 0.25384615384615383] 

Answer 1

從明確for循環，列表理解簡單地切換，並使用tuple代替list可以顯著加快你的操作：

from itertools import product 

ResultList = [(c, d, c[1]/d[1]) for a, b in product(t1, t2) for c, d in product(a, b)] 
#    ^Use a tuple here instead of a list 

如可以通過以下timeit試驗顯示：

>>> from timeit import Timer 
>>> original_test = Timer('original_version(Tuple1, Tuple2)', '''\ 
... Tuple1 = (
...  ((1, 2.2, 3), (2, 3.3, 4)), 
...  ((5, 6.6, 7), (6, 7.7, 8)) 
...) 
... Tuple2 = (
...  ((10, 11, 12), (11, 12, 13), (12, 13, 14)), 
...  ((20, 21, 22), (21, 22, 23), (22, 23, 24)) 
...) 
... def original_version(t1, t2): 
...  ResultList = [] 
...  for a in t1: 
...   for b in t2: 
...    for c in a: 
...     for d in b: 
...      ResultList.append([c, d, c[1]/d[1]])''') 
>>> improved_test = Timer('improved_version(Tuple1, Tuple2)', '''\ 
... from itertools import product 
... Tuple1 = (
...  ((1, 2.2, 3), (2, 3.3, 4)), 
...  ((5, 6.6, 7), (6, 7.7, 8)) 
...) 
... Tuple2 = (
...  ((10, 11, 12), (11, 12, 13), (12, 13, 14)), 
...  ((20, 21, 22), (21, 22, 23), (22, 23, 24)) 
...) 
... def improved_version(t1, t2): 
...  return [(c, d, c[1]/d[1]) for a, b in product(t1, t2) for c, d in product(a, b)]''') 
>>> original_time = original_test.timeit() 
>>> improved_time = improved_test.timeit() 
>>> print 'Improved version is %{} faster'.format(
...  (original_time - improved_time)/original_time * 100 
...) 
Improved version is %30.0181954314 faster 
>>>